Differentials — Question 5

Question 5 – Solution

Strategy. Compute component contributions, then confirm the relative formula by logarithmic structure.

Step 1: Differential $$dV=\frac{2\pi rh}{3}\,dr+\frac{\pi r^2}{3}\,dh.$$ At the given dimensions, $$dV=40\pi(0.04)+\frac{25\pi}{3}(-0.10) =1.6\pi-\frac 56\pi=\boxed{\frac{23\pi}{30}\ \mathrm{cm^3}}.$$

Step 2: Compare The radial contribution is $1.6\pi$; the height contribution is $-5\pi/6$. The radial contribution dominates, so the net predicted change is positive.

Step 3: Relative form $$\frac{dV}{V}=2\frac{dr}{r}+\frac{dh}{h} =2\frac{0.04}{5}-\frac{0.10}{12} =\boxed{\frac{23}{3000}}.$$ Also $V=100\pi$, and $(23\pi/30)/(100\pi)=23/3000$, verifying the result.